How many grams of Lead (II) Nitrate are required to make a 350mL solution of 2.5M Pb2+?-
Moles Pb = moles Pb(NO3)2.
How many moles Pb(NO3)2 do you need? moles = M x L = ??
moles = grams/molar mass
Solve for grams.
To find the number of grams of Lead (II) Nitrate required to make a 350 mL solution of 2.5 M Pb2+, you will need to follow a few steps:
Step 1: Determine the molar mass of Lead (II) Nitrate
Lead (II) Nitrate is composed of one lead ion (Pb2+) and two nitrate ions (NO3-).
Atomic masses:
- Pb: 207.2 g/mol
- N: 14.0 g/mol
- O: 16.0 g/mol
So, the molar mass of Lead (II) Nitrate (Pb(NO3)2) is:
(207.2 g/mol) + (2 * (14.0 g/mol + 3 * 16.0 g/mol)) = 207.2 g/mol + 166.0 g/mol = 373.2 g/mol
Step 2: Use the molar concentration (M) and the volume (V) to find the number of moles of Pb2+ ions.
Molar concentration (M) is defined as moles of solute per liter of solution. We have a 350 mL solution, which is equal to 0.35 L.
Number of moles (n) of solute (Pb2+) = Molarity (M) * Volume (V)
n = 2.5 mol/L * 0.35 L = 0.875 mol
Step 3: Calculate the mass of Lead (II) Nitrate required.
Mass (m) = Number of moles (n) * Molar mass (M)
m = 0.875 mol * 373.2 g/mol = 325.89 g
Therefore, approximately 325.89 grams of Lead (II) Nitrate are required to make a 350 mL solution of 2.5 M Pb2+.